Today I finished the proof for the bijectivity for the function f: (0,1)\to\mathbb{R} defined by f(x)=\frac{2x-1}{2x(1-x)}. Once we choose a real number $a$, we shall show f(x)=a has a unique solution in (0,1). This equation is equivalent to 2ax^2+(2-2a)x-1=0.

If a=0, there is only one solution x=1/2.

Otherwise a\neq 0, then we have two possible solutions, x_1 = \frac{a-1+\sqrt{a^2+1}}{2a} and x_2=\frac{a-1-\sqrt{a^2+1}}{2a}.

By the trick of multiplying conjugate of the numerator, for instance x_1=\frac{(a-1)^2-(a^2+1)}{2a(a-1-\sqrt{a^2+1})}=\frac{1}{1+(\sqrt{a^2+1}-a)}, we could have an estimate on x_1. Here, since \sqrt{a^2+1}-a > 0, x_1\in(0,1).

Similarly, one could show that x_2\notin(0,1).